0.596g of a gaseous compound containing only boron and hydrogen occupies 484 mL at NTP. When the compound was ignited in excess of oxygen, all its hydrogen was recovered as 1.17g of H_(2)O and all the boron was present as B_(2)O_(3). Find the molecular formula of the compound (B=10.8, H=1)

0.596g of a gaseous compound containing only boron and hydrogen occupi

1.	0.596g of a gaseous compound containing only boron and hydrogen occupies 484 mL at NTP. When the compound was ignited in excess of oxygen, all its hydrogen was recovered as 1.17g of H_(2)O and all the boron was present as B_(2)O_(3). Find the molecular formula of the compound (B=10.8, H=1)
Answer» Solution :484mL of the gaseous compound at NTP weighs 0.596g. `therefore` molecular weight of the compound =wt. of 1 mole of the compound =wt. of 22400mL at NTP `=(0.596)/(484)XX 22400=27.6` Let the FORMULA be `B_(x)H_(y)` The reaction may be written as, `{:(B_(x)H_(y) + O_(2) ,rarr ,H_(2)O + B_(2)O_(3)),(0.596g,,1.17g):}` APPLYING POAC for H atoms, `yxx` moles of `B_(x) H_(y)=2xx` moles of `H_(2)O` `y xx (0.596)/(27.6) = (2xx 1.17)/(18)` y=6 Further, for the formula `B_(x)H_(y)`, mol wt. =27.6 `therefore 10.8x + 1y= 27.6` `10.8x + 6= 27.6` x=2 Hence the molecular formula is `B_(2)H_(6)`